Answer
$-\dfrac{3}{5}$
Work Step by Step
Suppose $\theta =\tan^{-1} (\dfrac{3}{4})$
This gives: $\tan \theta=(-\dfrac{3}{4})$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ r=\sqrt {4^2+3^2}=5$
Therefore, we have $\sin \theta =\sin [\tan^{-1} (-\dfrac{3}{4})]=-\dfrac{3}{5}$
